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One of the major advantages of state space representation is the fact that it can represent non-LTI system as well. I tried searching a lot to find a reason but couldn't get a reason explicitly stating why tf method is not valid for non-LTI systems.

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  • \$\begingroup\$ How would you define the transfer function of a system that is not LTI? \$\endgroup\$ – Matt L. Oct 18 '14 at 11:30
  • \$\begingroup\$ That is precisely what my question is about. You're asking the same question to me. \$\endgroup\$ – Jay Oct 18 '14 at 11:51
  • \$\begingroup\$ Your question was why the transfer function method is not valid for non-LTI systems. My question was what you mean by transfer function if you talk about a non-LTI system. \$\endgroup\$ – Matt L. Oct 18 '14 at 13:03
  • \$\begingroup\$ If you wish you may see it in frequency domain too. Define TF as set of ratios of sine steady state output phasor over input one for each possible frequency. If your system is not LTI you don't get a steady sine output against sine stimulus, hence phasors and their ratioa are meaningless. A little more formally sine functions are not eigenvectors of non-LTI systems (TF would actually be eigenvalues as function of eigenvectors rapresentation) \$\endgroup\$ – carloc Nov 8 '16 at 10:34
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The reason why transfer functions work so well for linear time-invariant (LTI) systems (and don't for non-linear systems) is that they are the Laplace transform (or, in discrete time) the Z-transform of the system's impulse response, which completely characterizes the behavior of such systems. I.e., the impulse response, or, equivalently, the transfer function is all you need to know to compute the response of the system to any input signal. The reason for this is that any input signal can be written as an integral (or a sum in discrete time) of scaled and shifted delta impulses, and due to linearity and time invariance, the response to this integral/sum equals the integral/sum of scaled and shifted impulse responses:

$$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau\quad\Longrightarrow\quad y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{1}$$

where \$x(t)\$ is the input signal, \$y(t)\$ is the system's response, and \$h(t)\$ is the impulse response. Equation (1) is the convolution integral, which completely describes the system's behavior. In the transform domain (i.e., Laplace transform, Fourier transform, or Z-transform), convolution becomes multiplication.

For non-linear systems you can of course compute or measure the system's response to an impulse, but this function does not tell you anything about the system's response to other input signals, or even to a scaled impulse. This is why the impulse response and its transform do not have any significance for non-linear systems.

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A transfer function is some sort of transform of the unit impulse response of a system. Multiplying by a transfer function in the transform domain is the same as convolving in the time domain.

By definition, convolution cannot be used to predict the output of a nonlinear system. A nonlinear system does not have properties of additivity and scalability so you cannot break the input into pieces, calculate the output of each piece, and sum the outputs (i.e. convolution.

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  • \$\begingroup\$ How about a function that is linear but time variant? \$\endgroup\$ – Jay Oct 18 '14 at 16:06
  • \$\begingroup\$ Convolution does work for linear time varying systems. You just have a two-dimensional impulse response. See e.g. this. \$\endgroup\$ – Matt L. Oct 18 '14 at 19:09
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This is because the main concern is the system stability w.r.t time , i.e. if the input does not change with time, the same should be reflected in the output , hence the system should be time invariant.

Secondly, the system should be linear because in the theoretical analysis, the frequency/ B.w of the signal should not be altered by the system.

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Here's my intuitive understanding.

A transfer function tells you what your system's output looks like for many specific inputs. For instance, if you try the inputs \$\sin(t)\$ and \$\sin(2t)\$, maybe your outputs \$y(t)\$ are $$ y(\sin(t)) = 10\sin(t) $$ and $$ y(\sin(2t)) = 2 \sin(2t) $$ (This example is a low-pass filter - it amplifies lower frequencies more.)

Now, this is particularly useful in LTI systems for two reasons:

  1. If your input is made up of a bunch of these little parts, then you can use the transfer function to find the combined output. For instance, if your input was \$\sin(t) + \sin(2t)\$, then your output would be $$ y(\sin(t) + \sin(2t)) = y(\sin(t)) + y(\sin(2t)) = 10 \sin(t) + 2\sin(2t) $$ so the transfer function tells us how the system reacts to all types of inputs. This is not true for non-linear systems, so the transfer function isn't useful there.
  2. If your input is delayed (ie: \$\sin(t-1)\$), then the output is also delayed: $$ y(\sin(t - 1)) = 10\sin(t-1) $$ so our transfer function is valid for time-shifted inputs. Again, for time-variant systems, this isn't true, so if your input has phase shifts, then the transfer function is useless.

To sum up: LTI systems have nice properties that let us use a simple-looking transfer function to deal with non-simple inputs. When you take away linearity and time-invariance, the transfer function doesn't give you enough information to be helpful.

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